The average of the squares of the first 46 natural numbers is

Correct — D. You want the mean of the squares, not the square of the mean.

Sum of the first n squares = n(n+1)(2n+1)⁄6
Divide that by n: average = (n+1)(2n+1)⁄6 — the n cancels, so the sum is never needed.

Put n = 46:
(47 × 93)⁄6 = 4371⁄6 = 728.5 → option (d)

Check it backwards: 728.5 × 6 = 4371, and 47 × 93 = 4371.

Why the others are wrong

• (a)
  730.5 — 730.5 × 6 = 4383, not the 4371 that 47 × 93 gives. Multiplying an option back by 6 is the fastest check available, because the four choices sit exactly one apart and no estimate can separate them.
• (b)
  727.5 — 727.5 × 6 = 4365, six short of 4371. One unit of error in dividing 4371 by 6 lands you here, which is why the back-multiplication is worth four seconds.
• (c)
  729.5 — 729.5 × 6 = 4377, six over 4371. The mean of the first 46 squares is fixed by the single product 47 × 93, and 4377 is not that product.

Two different averages get confused here.

The sum of the first n squares is n(n+1)(2n+1)⁄6. Dividing by n gives the average, (n+1)(2n+1)⁄6.

The square of the average is a different number entirely: the mean of 1 to 46 is 23.5, and 23.5² = 552.25, nowhere near 728.5.

Averaging squares and squaring an average are not the same operation, and the gap between them grows with n.

Key facts

• Average of the squares of the first n natural numbers = (n+1)(2n+1)⁄6.
• For n = 46 that is (47 × 93)⁄6 = 4371⁄6 = 728.5.
• Average of the first n natural numbers = (n+1)⁄2, which for n = 46 is 23.5 and squares to 552.25.
• Sum of the first n cubes is [n(n+1)⁄2]², the square of the sum of the first n numbers.

Common traps

• Squaring the average, 23.5² = 552.25, instead of averaging the squares.
• Using the sum formula n(n+1)(2n+1)⁄6 and forgetting to divide by n.
• Slipping in the division 4371 ÷ 6 — the options are one apart, so a single slip picks a wrong one.

No directly related past PYQ was found.